# Laurent series and regions

Some help to understand why this is true would be amazing;

For the function $$\frac{1}{z-1}\;\;,$$ the laurent series is different depending on the region.

1) for the region $|z|<1$, you rewrite the function as $\frac{-1}{1-z}$, and the laurent series is $$-\sum_{n=0}^\infty z^n$$

2) for the region $1<|z|<2$, you rewrite the function as $\frac{1}{z}$ $\frac{1}{1-\frac{1}{z}}$, and the laurent series is $$\sum_{n=0}^\infty\frac{1}{z^{n+1}}$$

What I dont understand is, why can't I rewrite the functions in the same way, regardless of the region?

In both cases you use the geometric series expansion $$\frac{1}{1-z}=\sum_{k=0}^\infty z^n$$ which is only valid for $|z|<1$, so you cannot apply the same rule for $|z|>1$, unless you rewrite the expression as $$\frac{1}{z}\frac{1}{1-\frac{1}{z}}$$ in which case you can apply the geometric series expansion to the second factor because $|z|>1 \implies |1/z|<1$.
• I think I get it! So essentially, if the region is for example $|z|<1$, it implies that I should be working with a function that looks like $\frac{1}{1-z}$, and if I'm working with a region that looks like this $|z+1|>1 \implies |\frac{1}{z+1}|<1$, I should rewrite my function so that it looks like this $\frac{1}{1-\frac{1}{z+1}}$ . Am I correct? – armara Sep 29 '16 at 20:21
You can. The function is $\frac{1}{z-1}$. The point is that a Laurent series centered at some point (here $z=0$) is a power series expansion with both positive and negative integer powers. Such a series can only be convergent on some annuli bounded by singularities of the function. In the present case there is a singularity for $|z|=1$ which will then separate the plane into two annuli of convergence: $|z|<1$ (actually a disk) and $|z|>1$ (in fact, also a disk).